JEE Advanced (Single Correct Type): Mathematical induction & Binomial Theorem

Q.1. If xn – 1 is divisible by x – k, then the least positive integral value of k is
(a) 1
(b) 2
(c) 3
(d) 4

Given,
P(n): xn – 1 is divisible by x – k
Let us substitute n = 1, 2, 3,..
⇒ P(1) : x – 1
⇒ P(2) : x2 – 1 = (x−1)(x+1)
⇒ P(3) : x3 – 1 = (x − 1)(x+ x + 1)
⇒ P(4) : x4 − 1 = (x2 − 1)(x2 + 1) = (x − 1)(x + 1)(x2 + 1)
Therefore, the least positive integral value of k is 1.

Q.2. The coefficient of the middle term in the expansion of (2+3x)is:
(a) 5!
(b) 6
(c) 216
(d) 8!

If the exponent of the expression is n, then the total number of terms is n+1.
Hence, the total number of terms is 4+1 = 5.
Hence, the middle term is the 3rd term.
We know that general term of (x+a)n is Tr+1 =nCxn-r ar
Here, n=4, r=2
Therefore, T3 = 4C2.(2)2.(3x)2
T3 = (6).(4).(9x2)
T3 = 216x2.
Therefore, the coefficient of the middle term is 216.

Q.3. For any natural number n, 22n – 1 is divisible b
(a) 2
(b) 3
(c) 4
(d) 5

Let P(n) = 22n – 1
Substituting n = 1, 2, 3,….
P(1) = 22(1) – 1 = 4 – 1 = 3
This is divisible by 3.
P(2) = 22(2) – 1 = 16 – 1 = 15
This is divisible by 3.
P(3) = 22(3) – 1 = 256 – 1 = 255
This is also divisible by 3.
Assume that P(n) is true for some natural number k, i.e., P(k): 22k – 1 is divisible by 3, i.e., 22k – 1
= 3q, where q ∈ N
Now,
P(k + 1) : 22(k+1) – 1
= 22k + 2 – 1
= 22k . 22 – 1
= 22k . 4 – 1
= 3.22k + (22k – 1)
= 3.22k + 3q
= 3 (22k + q) = 3m, where m ∈ N
Thus P(k + 1) is true, whenever P(k) is true.
Therefore, for any natural number n, 22n – 1 is divisible by 3.

Q.4. The value of (126)1/3 up to three decimal places is
(a) 5.011
(b) 5.012
(c) 5.013
(d) 5.014

(126)⅓ can also be written as the cube root of 126.
Hence, (126) is approximately equal to 5.013.
Hence, option (c) 5.013 is the correct answer.

Q.5. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true
(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5

The student could be able to conclude that P(n) is true for all n ≥ 5 since P(5) is true for all k > 5 ∈ N as well as true for P(5) and P(k + 1) is true, whenever P(k) is true.

Q.6. If n is even in the expansion of (a+b)n, the middle term is:
(a) nth term
(b) (n/2)th term
(c) [(n/2)-1]th term
(d) [(n/2)+1]th term

In general, if “n” is the even in the expansion of (a+b)n, then the number of terms will be odd. (i.e) n+1.
Hence, the middle term of the expansion (a+b)n is [(n/2)+1]th term.

Q.7. If P (n): “49n + 16n + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is
(a) 1
(b) -2
(c) -1
(d) -3

Given that P(n) : 49n + 16n + k is divisible by 64 for n ∈ N
For n = 1,
P(1) : 49 + 16 + k = 65 + k is divisible by 64.
Thus k, should be -1 since, 65 – 1 = 64 is divisible by 64.

Q.8. The largest coefficient in the expansion of (1 + x)10 is:
(a) 10! / (5!)2
(b) 10! / 5!
(c) 10! / (5!×4!)2
(d) 10! / (5!×4!)

Given: (1+x)10
The greatest coefficient will always occur in the middle term.
Hence, the total number of terms in an expansion is 11. (i.e. 10+1 = 11)
Therefore, middle term = [(10/2) + 1] = 5+1 = 6th term.
We know that general term of (x+a)n is Tr+1 =nCxn-r ar
Here, n=10, r=5
So, T6 = 10C5.x5
Therefore, the coefficient of the greatest term = 10C5 = 10!/(5!)2.
So, option (a) 10!/(5!)2 is the correct answer.

Q.9. For all n ∈ N, 3.52n+1 + 23n+1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Let P(n) be the statement that 3.52n + 1 + 23n + 1 is divisible by 17
If n = 1, then given expression = 3 * 53 + 24 + 375 + 16 = 391 = 17 * 23, divisible by 17.
P(1) is true
Assume that P(k) is true.
3.52k + 1 + 23k + 1 is divisible by 17.
3.52k = 1 + 23k + 1 = 17m where m ∈ N
3.52(k + 1) + 1 + 23(k + 1) + 1
= 3.52k + 1 * 52 + 23k + 1 * 23
= 25(17m – 23k+1) + 8.23k + 1
= 425m – 25.23k + 1 + 8.23k + 1
= 425m – 17.23k + 1
= 17(25m – 23k + 1), divisible by 17
P(k + 1) is true by Principle of Mathematical Induction
P(n) is true for all n ∈ N. 3.52n + 1 + 23n + 1 is divisible by 17 for all n ∈ N

Q.10. The coefficient of x3y4 in (2x+3y2)5 is
(a) 360
(b) 720
(c) 240
(d) 1080

Given: (2x+3y2)5
Therefore, the general form for the expression (2x+3y2)5 is Tr+1 = 5Cr. (2x)r.(3y2)5-r
Hence, T3+1 = 5C3 (2x)3.(3y2)5-3
T4 = 5C3 (2x)3.(3y2)2
T4 = 5C3.8x3.9y4
On simplification, we get
T4 = 720x3y4
Therefore, the coefficient of x3y4 in (2x+3y2)5 is 720.

Q.11. n(n + 1) (n + 5) is a multiple of
(a) 3
(b) 8
(c) 5
(d) 7

Let P(n) = n(n + 1)(n + 5)
Substituting n = 1, 2, 3,….
P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6
P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7
P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12
..
Thus, from the above statements and verifying the options, we can say that n(n + 1)(n + 5) is a multiple of 3.

Q.12. The largest term in the expansion of (3+2x)50, when x = ⅕ is
(a) 6th term
(b) 7th term

(c) 8th term
(d) None of the above

The greatest term in the expansion of (x + y)n is the kth term. Where k = [(n + 1)y]/[x + y]..(1)
On comparing the given expression with the general form, x = 3, y=2x, n=50
Now, substitute the values in the given expression, we get
Hence, kth term = [(50 + 1)(2x)]/[3 + 2x]
When x =  ⅕,
Kth term = [(51)(2(⅕))]/[3 + 2(⅕)] = 6
Hence, the 6th term is the largest term in the expansion of (3 + 2x)50, when x = ⅕.

Q.13. n2 < 2n for all natural numbers
(a) n ≥ 5
(b) n < 5
(c) n > 1
(d) n ≤ 3

Consider, P(n) : n2 < 2n
Substituting n = 1, 2, 3,…
P(1): 12 < 21
1 < 2 (not true)
P(2): 22 < 22
4 < 4 (not true)
P(3): 32 < 23
9 < 8 (not true)
P(4): 42 < 24
16 < 16 (not true)
P(5): 52 < 25
25 < 32 (true)
P(6): 62 < 26
26 < 64 (true)
Thus, n2 < 2n for all natural numbers n ≥ 5.

Q.14. The coefficient of y in the expansion of (y2+(c/y))5 is:
(a) 10c
(b) 29c
(c) 10c3
(d) 20c3

Given: (y2+(c/y))5
We know that general term of (x+a)n is Tr+1 =nCxn-r ar
Here, n=5, r=?
(y2+(c/y))5 = 5Cr.(y2)r.(c/y)5-r
(y2+(c/y))5 = 5Cr. y2r. (c5-r/y5-r)
On solving this, we get r = 2.
Hence, the coefficient of y = 5C3.c3 = 10c3.
Therefore, option (c) 10c3 is the correct answer.

Q.15. If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is
(a) 5
(b) 3
(c) 7
(d) 1

Given that 10n + 3.4n+2 + k is exactly divisible by 9.
Consider: P(n) = 10n + 3.4n+2 + k
Substituting n = 1,
P(1) = 10+ 3.41+2 + k
= 10 + 3(64) + k
= 10 + 192 + k
= 202 + k is exactly divisible by 9, the value of k will be 5.

Q.16. The fourth term in the expansion of (x-2y)12 is:
(a) -1760 x9 × y3
(b) -1670 x9 × y3
(c) -7160 x9 × y3
(d) -1607 x9 × y3

We know that the general term of an expansion (a+b)n is Tr+1 = nCr an-r br.
Now, we have to find the fourth term in the expansion (x-2y)12
Hence, r = 3, a = x, b = -2y, n= 12.
Now, substitute the values in the formula, we get
T3+1 = 12C3 x12-3 (-2y)3.
On solving this, we get
T4 = -1760x9y3.

Q.17. Let P(n) : “2n < (1 × 2 × 3 × … × n)”. Then the smallest positive integer for which P(n) is true is
(a) 1
(b) 2
(c) 3
(d) 4

P(1) : 21 < 1
2< 1 is false
P(2) : 22 < 1 × 2
4 < 2 is false
P(3) : 23 < 1 × 2 × 3
8 < 6 is false
P(4) : 24 < 1 × 2 × 3 × 4
16 < 24 is true

Q.18. If the fourth term of the binomial expansion of (px+(1/x))n is 5/2, then
(a) n=6, p=6
(b) n=8, p=6
(c) n=8, p= ½
(d) n=6, p=½

Given: (px+(1/x))n
Hence, the fourth term, T3+1 = nC3(px)n-3(1/x)3
Given that the fourth term of the binomial expansion of (px+(1/x))n is 5/2, which is independent of x.
Hence, (5/2)= nC3(px)n-3(1/x)3 …(1)
On solving this, we get n=6.
Now, substitute n=6 in (5/2)= nC3(p)3
20p3 = 5/2
p3=⅛
p=½.
Therefore, n=6 and p=½.

Q.19. For every positive integer n, 7n – 3n is divisible by
(a) 3
(b) 4
(c) 7
(d) 5

Let P(n) = 7n – 3n
Substituting n = 1, 2, 3,…
P(1) = 71 – 31 = 7 – 3 = 4
P(2) = 72 – 32 = 49 – 9 = 40
P(3) = 73 – 33 = 343 – 27 = 316
Thus, for every positive integer n, 7n – 3n is divisible by 4.

Q.20. If n is the positive integer, then 23n – 7n -1 is divisible by
(a) 7
(b) 10
(c) 49
(d) 81

Given: 23n – 7n -1. It can also be written as 8n – 7n – 1
Let 8n – 7n – 1 =0
So, 8n = 7n+1
8n = (1+7)n
By applying binomial theorem, we get
8n – 1 – 7n = 49 (or) 23n – 7n -1 = 49
Hence, 23n – 7n -1 is divisible by 49.

The document JEE Advanced (Single Correct Type): Mathematical induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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